imone/ARB huggingface dataset:
Question: Let $\varphi(x, y)$ be a function with continuous second order partial derivatives such that

1. $\varphi_{x x}+\varphi_{y y}+\varphi_{x}=0$ in the punctured plane $\mathbb{R}^{2} \backslash\{0\}$,

2. $r \varphi_{x} \rightarrow \frac{x}{2 \pi r}$ and $r \varphi_{y} \rightarrow \frac{y}{2 \pi r}$ as $r=\sqrt{x^{2}+y^{2}} \rightarrow 0$.

Let $C_{R}$ be the circle $x^{2}+y^{2}=R^{2}$. Evaluate the line integral

$$
\int_{C_{R}} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)
$$


Answer: Consider the annular region $\mathcal{A}$ between the circles of radius $r$ and $R$, then by Green's Theorem

$$
\begin{aligned}
&\int_{R} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)-\int_{r} e^{y}\left(-\varphi_{y} d x+\varphi_{x} d y\right)= \\n&=\int_{\partial \mathcal{A}} e^{x}\left(-\varphi_{x} d x+\varphi_{x} d y\right) \\n&=\iint_{\mathcal{A}} d\left(e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)\right) \\n&=\iint_{\mathcal{A}}-e^{x} \varphi_{y y} d y \wedge d x+\left(e^{x} \varphi_{x}+e^{x} \varphi_{x x}\right) d x \wedge d y \\n&=\iint_{\mathcal{A}} e^{x}\left(\varphi_{x x}+e^{x} \varphi_{x x}+\varphi_{x}\right) d x \wedge d y=0
\end{aligned}
$$

showing that the integral does not depend on the radius $r$. Now, parametrizing the circle of radius $r$

$$
\begin{aligned}
&\int_{C_{r}} e^{x}\left(-\varphi_{y} d x+\varphi_{x} d y\right)= \\n&=\int_{0}^{2 \pi} e^{r \cos \theta}\left(\varphi_{y}(r \cos \theta, r \sin \theta) \sin \theta+\varphi_{x}(r \cos \theta, r \sin \theta) \cos \theta\right) r d \theta
\end{aligned}
$$

but when $r \rightarrow 0$ the integrand converges uniformly to

$$
\frac{\sin \theta}{2 \pi} \cdot \sin \theta+\frac{\cos \theta}{2 \pi} \cdot \cos \theta=\frac{1}{2 \pi}
$$

so the integral approaches $1$ when $r \rightarrow 0$ and that is the value of the integral.
Question: Compute$$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1} . $$
Answer: We use the fact that$$ \lim _{x \rightarrow 0^{+}} x^{x}=1 . $$As a consequence, we have$$ \lim _{x \rightarrow 0^{+}} \frac{x^{x+1}}{x}=1 . $$For our problem, let $\epsilon>0$ be a fixed small positive number. There exists $n(\epsilon)$ such that for any integer $n \geq n(\epsilon)$,$$ 1-\epsilon<\frac{\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}}{\frac{k}{n^{2}}}<1+\epsilon, \quad k=1,2, \ldots, n . $$From this, using properties of ratios, we obtain$$ 1-\epsilon<\frac{\sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}}{\sum_{k=1}^{n} \frac{k}{n^{2}}}<1+\epsilon, \quad \text { for } n \geq n(\epsilon) . $$Knowing that $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$, this implies$$ (1-\epsilon) \frac{n+1}{2 n}<\sum_{k=1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}<(1+\epsilon) \frac{n+1}{2 n}, \quad \text { for } n \geq n(\epsilon) . $$It follows that$$ \lim _{n \rightarrow \infty} \sum_{k-1}^{n}\left(\frac{k}{n^{2}}\right)^{\frac{k}{n^{2}}+1}=\frac{1}{2} . $$
Question: A sequence $\left\{x_{n}, n \geq 1\right\}$ satisfies $x_{n+1}=x_{n}+e^{-x_{n}}, n \geq 1$, and $x_{1}=1$. Find $\lim _{n \rightarrow \infty} \frac{x_{n}}{\ln n}$.
Answer: The sequence $\left\{x_{n}\right\}$ is increasing, hence it has a finite or infinite limit. If $\lim _{n \rightarrow \infty} x_{n}=$ $x<+\infty$, then $x=x+e^{-x}$, a contradiction. Therefore, $\lim _{n \rightarrow \infty} x_{n}=+\infty$. We use twice the Stolz-Cesaro theorem and obtain  $$ \lim _{n \rightarrow \infty} \frac{x_{n}}{\ln n}=\lim _{n \rightarrow \infty} \frac{x_{n+1}-x_{n}}{\ln (n+1)-\ln n}=\lim _{n \rightarrow \infty} \frac{e^{-x_{n}}}{\ln \left(1+\frac{1}{n}\right)}=\lim _{n \rightarrow \infty} \frac{n}{e^{x_{n}}}=\lim _{n \rightarrow \infty} \frac{1}{e^{x_{n+1}}-e^{x_{n}}} . $$  It remains to notice that  $$ \lim _{n \rightarrow \infty}\left(e^{x_{n+1}}-e^{x_{n}}\right)=\lim _{n \rightarrow \infty}\left(e^{x_{n}+e^{-x_{n}}}-e^{x_{n}}\right)=\lim _{n \rightarrow \infty} \frac{e^{e^{-x_{n}}}-1}{e^{-x_{n}}}=\lim _{t \rightarrow 0} \frac{e^{t}-1}{t}=1 . $$  Answer: 1
Question: Find the limit  $$ \lim _{N \rightarrow \infty} \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right), $$  where $\{x\}$ denotes the fractional part of $x$.
Answer: Estimate $\max _{1 \leq n \leq N}\{\sqrt{n}\}$. For $k^{2} \leq n<(k+1)^{2}$ it holds  $$ \{\sqrt{n}\}=\sqrt{n}-k \leq \sqrt{(k+1)^{2}-1}-k, $$  where equality is attained for $n=(k+1)^{2}-1$. If $k<l$ then  $$ \begin{aligned} 1 &-\left\{\sqrt{(k+1)^{2}-1}\right\}=k+1-\sqrt{(k+1)^{2}-1}=\\ &=\frac{1}{k+1+\sqrt{(k+1)^{2}-1}}>\frac{1}{l+1+\sqrt{(l+1)^{2}-1}}=1-\left\{\sqrt{(l+1)^{2}-1}\right\}, \end{aligned} $$  hence $\left\{\sqrt{(k+1)^{2}-1}\right\}$ is increasing in $k$. Thus, for $N=m^{2}-1$ we get  $$ \max _{1 \leq n \leq N}\{\sqrt{n}\}=\left\{\sqrt{m^{2}-1}\right\}=\sqrt{m^{2}-1}-m+1 . $$  Now consider $m^{2} \leq N \leq(m+1)^{2}-1$. Since $\max _{1 \leq n \leq N}\{\sqrt{n}\}$ is increasing in $N$, it holds  $$ \sqrt{m^{2}-1}-m+1 \leq \max _{1 \leq n \leq N}\{\sqrt{n}\} \leq \sqrt{(m+1)^{2}-1}-m . $$  Therefore,  $$ 1+m-\sqrt{(m+1)^{2}-1} \leq 1-\max _{1 \leq n \leq N}\{\sqrt{n}\} \leq m-\sqrt{m^{2}-1} $$  Also we have $m \leq \sqrt{N}<m+1$, so  $$ \begin{aligned} &\frac{m}{m+1} \cdot(m+1)\left(m+1-\sqrt{(m+1)^{2}-1}\right) \leq \\ &\quad \leq \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right)<\frac{m+1}{m} \cdot m\left(m-\sqrt{m^{2}-1}\right) . \end{aligned} $$  Since $m\left(m-\sqrt{m^{2}-1}\right)=\frac{m}{m+\sqrt{m^{2}-1}} \rightarrow \frac{1}{2}$ and $\frac{m}{m+1} \rightarrow 1$, as $m \rightarrow \infty$, we have $\lim _{N \rightarrow \infty} \sqrt{N}\left(1-\max _{1 \leq n \leq N}\{\sqrt{n}\}\right)=\frac{1}{2}$  Answer: $\frac{1}{2}
Question: Compute

$$
L=\lim _{n \rightarrow \infty}\left(\frac{n^{n}}{n !}\right)^{1 / n} .
$$


Answer:  Let $p_{1}=1, p_{2}=(2 / 1)^{2}, p_{3}=(3 / 2)^{3}, \ldots, p_{n}=(n /(n-1))^{n}$. Then

$$
\frac{p_{1} p_{2} \cdots p_{n}}{n}=\frac{n^{n}}{n !},
$$

and since $p_{n} \rightarrow e$, we have $\lim \left(n^{n} / n !\right)^{1 / n}=e$ as well (using the fact that $\left.\lim n^{1 / n}=1\right)$.
Question: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be the function of period $2 \pi$ such that $f(x)=x^{3}$ for $-\pi \leqslant x<\pi$.

The Fourier series for $f$ has the form $\sum_{1}^{\infty} b_{n} \sin n x$. Find
$$
\sum_{n=1}^{\infty} b_{n}^{2}.
$$


Answer: Using Parseval's Identity 

$$
\frac{1}{2} a_{0}^{2}+\sum_{n=1}^{\infty}\left(a_{n}^{2}+b_{n}^{2}\right)=\frac{1}{\pi} \int_{-\pi}^{\pi} f^{2}(x) d x
$$

and the fact that all $a_{n}=0$,

$$
\sum_{n=1}^{\infty} b^{2}=\frac{1}{\pi} \int_{-\pi}^{\pi} x^{6} d x=\frac{2}{7} \pi^{6} .
$$
Question: For a real $2 \times 2$ matrix

$$
X=\left(\begin{array}{ll}
 x & y \\
 z & t
\end{array}\right),
$$

let $\|X\|=x^{2}+y^{2}+z^{2}+t^{2}$, and define a metric by $d(X, Y)=\|X-Y\|$. Let $\Sigma=\{X \mid \operatorname{det}(X)=0\}$. Let

$$
A=\left(\begin{array}{ll}
 1 & 0 \\
 0 & 2
\end{array}\right) \text {. }
$$

Find the minimum distance from $A$ to $\Sigma$.
Answer: For $X \in \Sigma$, we have

$$
\begin{aligned}
\|A-X\|^{2} &=(1-x)^{2}+y^{2}+z^{2}+(2-t)^{2} \\n&=y^{2}+z^{2}+1-2 x+x^{2}+(2-t)^{2} \\n& \geqslant \pm 2 y z+1-2 x+2|x|(2-t) \\n&=4|x|-2 x+2(\pm y z-|x| t)+1
\end{aligned}
$$

We can choose the sign, so $\pm y z-|x| t=0$ because $\operatorname{det} X=0$. As $4|x|-2 x \geqslant 0$, we have $\|A-X\| \geqslant 1$ with equality when $4|x|-2 x=0,|x|=2-t, y=\pm z$,
Question: Find the probability that in the process of repeatedly flipping an unbiased coin, one will encounter a run of 5 heads before one encounters a run of 2 tails.
Answer: We call a successful string a sequence of $H$ 's and $T$ 's in which $H H H H H$ appears before $T T$ does. Each successful string must belong to one of the following three types:(i) those that begin with $T$, followed by a successful string that begins with $H$;(ii) those that begin with $H, H H, H H H$, or $H H H H$, followed by a successful string that begins with $T$;(iii) the string $H H H H H$.Let $P_{H}$ denote the probability of obtaining a successful string that begins with $H$, and let $P_{T}$ denote the probability of obtaining a successful string that begins with $T$. Then$$ \begin{aligned} P_{T} &=\frac{1}{2} P_{H}, \\ P_{H} &=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right) P_{T}+\frac{1}{32} . \end{aligned} $$Solving these equations simultaneously, we find that$$ P_{H}=\frac{1}{17} \quad \text { and } \quad P_{T}=\frac{1}{34} . $$Hence the probability of obtaining five heads before obtaining two tails is $\frac{3}{34}$.
Question: Compute the integral $\iint_{D} x d x d y$, where$$ D=\left\{(x, y) \in \mathbb{R}^{2} \mid x \geq 0,1 \leq x y \leq 2,1 \leq \frac{y}{x} \leq 2\right\} . $$
Answer: The domain is bounded by the hyperbolas $x y=1, x y=2$ and the lines $y=x$ and $y=2 x$. This domain can mapped into a rectangle by the transformation$$ T: \quad u=x y, \quad v=\frac{y}{x} . $$Thus it is natural to consider the change of coordinates$$ T^{-1}: \quad x=\sqrt{\frac{u}{v}}, \quad y=\sqrt{u v} . $$The domain becomes the rectangle $D^{*}=\left\{(u, v) \in \mathbb{R}^{2} \mid 1 \leq u \leq 2,1 \leq v \leq 2\right\}$. The Jacobian of $T^{-1}$ is $\frac{1}{2 v} \neq 0$. The integral becomes$$ \int_{1}^{2} \int_{1}^{2} \sqrt{\frac{u}{v}} \frac{1}{2 v} d u d v=\frac{1}{2} \int_{1}^{2} u^{1 / 2} d u \int_{1}^{2} v^{-3 / 2} d v=\frac{1}{3}(5 \sqrt{2}-6) . $$
Question: A husband and wife agree to meet at a street corner between 4 and 5 o'clock to go shopping together. The one who arrives first will await the other for 15 minutes, and then leave. What is the probability that the two meet within the given time interval, assuming that they can arrive at any time with the same probability? 
Answer: Denote by $x$, respectively, $y$, the fraction of the hour when the husband, respectively, wife, arrive. The configuration space is the square $$ D=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} . $$In order for the two people to meet, their arrival time must lie inside the region$$ D_{f}=\left\{(x, y)|| x-y \mid \leq \frac{1}{4}\right\} . $$The desired probability is the ratio of the area of this region to the area of the square.The complement of the region consists of two isosceles right triangles with legs equal to $\frac{3}{4}$, and hence of areas $\frac{1}{2}\left(\frac{3}{4}\right)^{2}$. We obtain for the desired probability$$ 1-2 \cdot \frac{1}{2} \cdot\left(\frac{3}{4}\right)^{2}=\frac{7}{16} \approx 0.44 . $$
Question: Evaluate the product$$ \left(1-\cot 1^{\circ}\right)\left(1-\cot 2^{\circ}\right) \cdots\left(1-\cot 44^{\circ}\right) \text {. } $$
Answer: We have$$ \begin{aligned} \left(1-\cot 1^{\circ}\right)\left(1-\cot 2^{\circ}\right) \cdots\left(1-\cot 44^{\circ}\right) \\ &=\left(1-\frac{\cos 1^{\circ}}{\sin 1^{\circ}}\right)\left(1-\frac{\cos 2^{\circ}}{\sin 2^{\circ}}\right) \cdots\left(1-\frac{\cos 44^{\circ}}{\sin 44^{\circ}}\right) \\ &=\frac{\left(\sin 1^{\circ}-\cos 1^{\circ}\right)\left(\sin 2^{\circ}-\cos 2^{\circ}\right) \cdots\left(\sin 44^{\circ}-\cos 44^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 44^{\circ}} \end{aligned} $$Using the identity $\sin a-\cos a=\sqrt{2} \sin \left(a-45^{\circ}\right)$ in the numerators, we transform this further into$$ \begin{gathered} \frac{\sqrt{2} \sin \left(1^{\circ}-45^{\circ}\right) \cdot \sqrt{2} \sin \left(2^{\circ}-45^{\circ}\right) \cdots \sqrt{2} \sin \left(44^{\circ}-45^{\circ}\right)}{\sin 1^{\circ} \sin 2^{\circ} \cdots \sin 44^{\circ}} \\ =\frac{(\sqrt{2})^{44}(-1)^{44} \sin 44^{\circ} \sin 43^{\circ} \cdots \sin 1^{\circ}}{\sin 44^{\circ} \sin 43^{\circ} \cdots \sin 1^{\circ}} . \end{gathered} $$After cancellations, we obtain $2^{22}$.
Question: Find the number of roots of

$$
z^{7}-4 z^{3}-11=0
$$

which lie between the two circles $|z|=1$ and $|z|=2$.


Answer:  Let $p(z)=z^{7}-4 z^{3}-11$. For $z$ in the unit circle, we have

$$
|p(z)-11|=\left|z^{7}-4 z^{3}\right| \leqslant 5<11
$$

so, by Rouch\u00e9's Theorem, the given polynomial has no zeros in the unit disc. For $|z|=2$,

$$
\left|p(z)-z^{7}\right|=\left|4 z^{3}+11\right| \leqslant 43<128=\left|z^{7}\right|
$$

so there are seven zeros inside the disc $\{z|| z \mid<2\}$ and they are all between the two given circles.
Question: Let $\mathcal{S}=\left\{(x, y, z) \in \mathbb{R}^{3} \mid x^{2}+y^{2}+z^{2}=1\right\}$ denote the unit sphere in $\mathbb{R}^{3}$. Evaluate the surface integral over $\mathcal{S}$ :

$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A .
$$


Answer:  Using the change of variables

$$
\begin{cases}x=\sin \varphi \cos \theta & 0<\theta<2 \pi \\ y=\sin \varphi \sin \theta & 0<\varphi<\pi \\ z=\cos \varphi & \end{cases}
$$

we have

$$
d A=\sin \varphi d \theta d \varphi
$$

and

$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A=\int_{0}^{\pi} \int_{0}^{2 \pi}\left(\sin ^{2} \varphi \cos ^{2} \theta+\sin \varphi \sin \theta+\cos \varphi\right) \sin \varphi d \theta d \varphi .
$$

Breaking the integral in three terms, we get

$$
\begin{aligned}
\int_{0}^{\pi} \int_{0}^{2 \pi} \sin \varphi \cos \varphi d \theta d \varphi=2 \pi \cdot \frac{1}{2} \int_{0}^{\pi} \sin 2 \varphi d \varphi=0 \\n\int_{0}^{\pi} \int_{0}^{2 \pi} \sin ^{2} \varphi \sin \theta d \theta d \varphi=\left(\int_{0}^{\pi} \sin ^{2} \varphi d \varphi\right) \int_{0}^{2 \pi} \sin \theta d \theta=0 \\n\int_{0}^{\pi} \int_{0}^{2 \pi} \sin ^{3} \varphi \cos ^{2} \theta d \theta d \varphi &=\left(\int_{0}^{\pi} \sin ^{3} \varphi d \varphi\right)\left(\int_{0}^{2 \pi} \cos ^{2} \theta d \theta\right) \\n&=\int_{0}^{\pi} \frac{1}{4}\left(3 \sin \varphi-\sin ^{3} \varphi\right) d \varphi \int_{0}^{2 \pi} \cos ^{2} \theta d \theta \\n&=\frac{1}{4}\left(-3 \cos \varphi+\left.\frac{1}{3} \cos ^{3} \varphi\right|_{0} ^{\pi}\right)\left(\int_{0}^{2 \pi} \frac{1+\cos 2 \theta}{2} d \theta\right) \\n&=\frac{1}{4}\left(-3(-2)+\frac{1}{3}(-2)\right) \pi \\n&=\frac{1}{4}\left(6-\frac{2}{3}\right) \pi=\frac{4}{3} \pi .
\end{aligned}
$$

Therefore,

$$
\iint_{\mathcal{S}}\left(x^{2}+y+z\right) d A=\frac{4}{3} \pi
$$
Question: Find the sum of the series  $$ \sum_{n=0}^{\infty} \frac{1}{n ! 2^{n}} \cos \frac{\pi n-1}{2} . $$
Answer: For all $x \in \mathbb{R}$, it holds  $$ \begin{aligned} &\sum_{n=0}^{\infty} \frac{\cos \left(\frac{\pi n}{2}-x\right)}{n !} x^{n}=\cos x \sum_{n=0}^{\infty} \frac{\cos \frac{\pi n}{2}}{n !} x^{n}+\sin x \sum_{n=0}^{\infty} \frac{\sin \frac{\pi n}{2}}{n !} x^{n}= \\ &=\cos x \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}+\sin x \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1) !} x^{2 k+1}=\cos ^{2} x+\sin ^{2} x=1 . \end{aligned} $$  In particular for $x=\frac{1}{2}$ we obtain the sum of the series in question.  Answer: 1
Question: What is the probability that a uniformly random permutation of the first $n$ positive integers has the numbers 1 and 2 within the same cycle?
Answer: The total number of permutations is of course $n$ !. We will count instead the number of permutations for which 1 and 2 lie in different cycles.If the cycle that contains 1 has length $k$, we can choose the other $k-1$ elements in $\left(\begin{array}{c}n-2 \\ k-1\end{array}\right)$ ways from the set $\{3,4, \ldots, n\}$. There exist $(k-1)$ ! circular permutations of these elements, and $(n-k)$ ! permutations of the remaining $n-k$ elements. Hence the total number of permutations for which 1 and 2 belong to different cycles is equal to$$ \sum_{k=1}^{n-1}\left(\begin{array}{l} n-2 \\ k-1 \end{array}\right)(k-1) !(n-k) !=(n-2) ! \sum_{k=1}^{n-1}(n-k)=(n-2) ! \frac{n(n-1)}{2}=\frac{n !}{2} . $$It follows that exactly half of all permutations contain 1 and 2 in different cycles, and so half contain 1 and 2 in the same cycle. The probability is $\frac{1}{2}$.
Question: Find$$ \int_{0}^{1} \frac{\ln (1+x)}{1+x^{2}} d x .$$
Answer: With the substitution $\arctan x=t$ the integral takes the form$$ I=\int_{0}^{\frac{\pi}{4}} \ln (1+\tan t) d t . $$This we already computed in the previous problem. ("Happiness is longing for repetition," says M. Kundera.) So the answer to the problem is $\frac{\pi}{8} \ln 2$.
Question: Let $C$ be the unit circle $x^{2}+y^{2}=1$. A point $p$ is chosen randomly on the circumference of $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$-and $y$-axes with diagonal $p q$. What is the probability that no point of $R$ lies outside of $C$?
Answer: The pair $(p, q)$ is chosen randomly from the three-dimensional domain $C \times$ int $C$, which has a total volume of $2 \pi^{2}$ (here int $C$ denotes the interior of $C$ ). For a fixed $p$, the locus of points $q$ for which $R$ does not have points outside of $C$ is the rectangle whose diagonal is the diameter through $p$ and whose sides are parallel to the coordinate axes (Figure 112). If the coordinates of $p$ are $(\cos \theta, \sin \theta)$, then the area of the rectangle is $2|\sin 2 \theta|$Figure 112The volume of the favorable region is therefore$$ V=\int_{0}^{2 \pi} 2|\sin 2 \theta| d \theta=4 \int_{0}^{\pi / 2} 2 \sin 2 \theta d \theta=8 . $$Hence the probability is$$ P=\frac{8}{2 \pi^{2}}=\frac{4}{\pi^{2}} \approx 0.405 . $$
Question: Let $M$ be a subset of $\{1,2,3, \ldots, 15\}$ such that the product of any three distinct elements of $M$ is not a square. Determine the maximum number of elements in $M$.
Answer: Note that the product of the three elements in each of the sets $\{1,4,9\},\{2,6,12\}$, $\{3,5,15\}$, and $\{7,8,14\}$ is a square. Hence none of these sets is a subset of $M$. Because they are disjoint, it follows that $M$ has at most $15-4=11$ elements.Since 10 is not an element of the aforementioned sets, if $10 \notin M$, then $M$ has at most 10 elements. Suppose $10 \in M$. Then none of $\{2,5\},\{6,15\},\{1,4,9\}$, and $\{7,8,14\}$ is a subset of $M$. If $\{3,12\} \not \subset M$, it follows again that $M$ has at most 10 elements. If $\{3,12\} \subset M$, then none of $\{1\},\{4\},\{9\},\{2,6\},\{5,15\}$, and $\{7,8,14\}$ is a subset of $M$, and then $M$ has at most 9 elements. We conclude that $M$ has at most 10 elements in any case.Finally, it is easy to verify that the subset$$ M=\{1,4,5,6,7,10,11,12,13,14\} $$has the desired property. Hence the maximum number of elements in $M$ is 10 .
Question: Compute the limit 

$$
\lim _{z \rightarrow 0}\left((\tan z)^{-2}-z^{-2}\right)
$$

 where $z$ is a complex variable.


Answer:  As

$$
(\tan z)^{-2}-z^{-2}=\frac{z^{2}-(\tan z)^{2}}{z^{2}(\tan z)^{2}}
$$

the Maclaurin expansion of the numerator has no terms of degree up to 3 , whereas the expansion of the denominator starts with $z^{4}$, therefore, the limit is finite.

As

$$
\tan z=z+\frac{1}{3} z^{3}+o\left(z^{4}\right) \quad(z \rightarrow 0)
$$

we have

$$
(\tan z)^{-2}-z^{-2}=\frac{z^{2}-z^{2}-\frac{2}{3} z^{4}+o\left(z^{4}\right)}{z^{4}+o\left(z^{4}\right)} \quad(z \rightarrow 0)
$$

so the limit at 0 is $-2 / 3$.
Question: Let $f(x)=\frac{1}{4}+x-x^{2}$. For any real number $x$, define a sequence $\left(x_{n}\right)$ by $x_{0}=x$ and $x_{n+1}=f\left(x_{n}\right)$. If the sequence converges, let $x_{\infty}$ denote the limit. For $x=0$, find $x_{\infty}=\lambda$.


Answer: We have

$$
f(x)=\frac{1}{2}-\left(x-\frac{1}{2}\right)^{2}
$$

so $x_{n}$ is bounded by $1 / 2$ and, by the Induction Principle, nondecreasing. Let $\lambda$ be its limit. Then

$$
\lambda=\frac{1}{2}-\left(\lambda-\frac{1}{2}\right)^{2}
$$

and, as the sequence takes only positive values,

$$
\lambda=\frac{1}{2} \text {. }
$$
Question: The clock-face is a disk of radius 1 . The hour-hand is a disk of radius $1 / 2$ is internally tangent to the circle of the clock-face, and the minute-hand is a line segment of length 1 . Find the area of the figure formed by all intersections of the hands in 12 hours (i.e., in one full turn of the hour-hand).
Answer: Let the minute hand makes an angle $\varphi \in[0 ; 2 \pi)$ with the vertical position. At that moment, the hour hand makes one of the angles  $$ \psi_{k}=\frac{\varphi+2 \pi k}{12}(\bmod 2 \pi), k \in \mathbb{Z}, $$  with the vertical position. The hands intersect along a segment of length  $$ \max \left(0, \cos \left(\psi_{k}-\varphi\right)\right), $$  hence the area in question equals $\frac{1}{2} \int_{0}^{2 \pi} \rho^{2}(\varphi) d \varphi$, where  $$ \rho(\varphi)=\max \left(0, \max _{k} \cos \left(\psi_{k}-\varphi\right)\right), \varphi \in \mathbb{R} . $$  Notice that  $$ \rho(\varphi)=\cos \left(\min _{k}\left|\frac{\varphi+2 \pi k}{12}-\varphi\right|\right)=\cos \left(\min _{k}\left|\frac{11 \varphi}{12}-\frac{\pi k}{6}\right|\right) . $$  This function is periodic with period of $\frac{2 \pi}{11}$. Since for $\varphi \in\left[-\frac{\pi}{11}, \frac{\pi}{11}\right]$ it holds $\rho(\varphi)=$ $\cos \left(\frac{11 \varphi}{12}\right)$, the area of the figure formed by intersections of hands equals  $$ \begin{gathered} \frac{1}{2} \int_{0}^{2 \pi} \rho^{2}(\varphi) d \varphi=\frac{11}{2} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}} \rho^{2}(\varphi) d \varphi=\frac{11}{2} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}} \cos ^{2}\left(\frac{11 \varphi}{12}\right) d \varphi= \\ =\frac{11}{4} \int_{-\frac{\pi}{11}}^{\frac{\pi}{11}}\left(1+\cos \left(\frac{11 \varphi}{6}\right)\right) d \varphi=\left.\left(\frac{11 \varphi}{4}+\frac{3}{2} \sin \left(\frac{11 \varphi}{6}\right)\right)\right|_{-\frac{\pi}{11}} ^{\frac{\pi}{11}}=\frac{\pi+3}{2} . \end{gathered} $$  Answer: $\frac{\pi+3}{2}
Question: Evaluate

$$
\iint_{\mathcal{A}} e^{-x^{2}-y^{2}} d x d y,
$$

where $\mathcal{A}=\left\{(x, y) \in \mathbb{R}^{2} \mid x^{2}+y^{2} \leqslant 1\right\}$.


Answer:  Using polar coordinates, we have

$$
\begin{aligned}
\iint_{\mathcal{A}} e^{-x^{2}-y^{2}} d x d y &=\int_{0}^{2 \pi} \int_{0}^{1} \rho e^{-\rho^{2}} d \rho d \theta \\n&=-\frac{1}{2} \int_{0}^{2 \pi} \int_{0}^{1}-2 \rho e^{-\rho^{2}} d \rho d \theta \\n&=-\frac{1}{2} \int_{0}^{2 \pi}\left(e^{-1}-1\right) \\n&=\pi\left(e^{-1}-1\right) .
\end{aligned}
$$
Question: Let $S_{9}$ denote the group of permutations of nine objects and let $A_{9}$ be the subgroup consisting of all even permutations. Denote by $1 \in S_{9}$ the identity permutation. Determine the minimum of all positive integers $m$ such that every $\sigma \in A_{9}$ satisfies $\sigma^{m}=1$.


Answer:  The order of a $k$-cycle is $k$, so the smallest $m$ which simultaneously annihilates all 9-cycles, 8-cycles, 7-cycles, and 5-cycles is $2^{3} \cdot 3^{2} \cdot 5 \cdot 7= 2520$. Any $n$-cycle, $n \leqslant 9$, raised to this power is annihilated, so $n=2520$.

To compute $n$ for $A 9$, note that an 8-cycle is an odd permutation, so no 8-cycles are in $A_{9}$. Therefore, $n$ need only annihilate 4-cycles (since a 4-cycle times a transposition is in A9), 9-cycles, 7-cycles, and 5-cycles. Thus, $n=2520 / 2= 1260$.
Question: A bag contains 1993 red balls and 1993 black balls. We remove two uniformly random balls at a time repeatedly and

(i) discard them if they are of the same color,

(ii) discard the black ball and return to the bag the red ball if they are of different colors.

The process ends when there are less than two balls in the bag. What is the probability that this process will terminate with one red ball in the bag?
Answer: First, observe that since at least one ball is removed during each stage, the process will eventually terminate, leaving no ball or one ball in the bag. Because red balls are removed 2 at a time and since we start with an odd number of red balls, the number of red balls in the bag at any time is odd. Hence the process will always leave red balls in the bag, and so it must terminate with exactly one red ball. The probability we are computing is therefore 1 .